Points \(P\) and \(Q\) lie on side \(BC\) of acute-angled triangle \(ABC\) so that \( \angle PAB = \angle BCA \) and \( \angle CAQ = \angle ABC\). Points \(M\) and \(N\) lie on lines \(AP\) and \(AQ\) respectively, such that \(P\) is the midpoint of \(AM\), and \(Q\) is the midpoint of \(AN\). Prove that lines \(BM\) and \(CN\) intersect on the circumcircle of triangle \(ABC\).

## Comments

Sort by:

TopNewestI'll give a solution using complex numbers, hopefully no mistakes were made. The triangles QCA and ACB are similar, so (C-Q)/(A-Q)=[(C-A)/(B-A)]£, where £ refers to complex conjugation. Also, triangles ABC, PBA are similar, so (A-P)/(B-P)=[(C-A)/(B-A)]£ . But it is obvious that M-P=P-A and N-Q=Q-A, using these relations we get that (C-Q)/(N-Q)=(M-P)/(B-P), which means that the triangles CQN and MPB are similar. Now we observe that angle(ABM) +angle(ACN)=180 degrees, which means that ABCD is cyclic, so obviously D is on the circumcircle of the triangle ABC. – Adrian Stefan · 2 years, 3 months ago

Log in to reply

– Sreejato Bhattacharya · 2 years, 3 months ago

Alternatively, you could note that \(\triangle BPM \sim \triangle CQN\) and then perform a not too tedious trig bash.Log in to reply

– Zi Song Yeoh · 2 years, 3 months ago

My friend said that he solved it with Barycentric coordinates in 5 minutes :OLog in to reply